In the problem 62 × 45, what are the partial products?

Mathematics

1 answer:

Nana76 [90]3 years ago

4 0

In the problem 62 times 45 the partial numbers are 6 and 5

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a boat costs 11,850 and decreases in value by 10% per year. how much will the boat be worth after 8 years

lawyer [7]

If the boat initially cost $11850 and INCREASED in value by 10% per year, we'd calculate the final value like this:

V = $11850(1+0.10)^8

But if the boat is losing value (depreciating) over time, we'd calculate its final value like this:

V = $11850(1-0.10)^8 = $11850(0.90)^8 = $11850(0.4305) = $5101.04 (answer)

5 0

3 years ago

Directions for questions 4 & 5: We selected a random sample of 100 StatCrunchU students, 67 females and 33 males, and analyz

GaryK [48]

Answer:

The 95% confidence interval for the difference between means is (-2164.21, -299.13).

The lower limit on the confidence interval is -$2164.21.

The upper limit on the confidence interval is -$299.13.

Step-by-step explanation:

The sample data is:

Gender Mean Std. dev. n

Female 2577.75 1916.29 67

Male 3809.42 2379.47 33

We have to calculate a 95% confidence interval for the difference between means, with a T-model.

The sample 1, of size n1=67 has a mean of 2577.75 and a standard deviation of 1916.29.

The sample 2, of size n2=33 has a mean of 3809.42 and a standard deviation of 2379.47.

The difference between sample means is Md=-1231.67.

$M_d=M_1-M_2=2577.75-3809.42=-1231.67$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{1916.29^2}{67}+\dfrac{2379.47^2}{33}}\\\\\\s_{M_d}=\sqrt{54808.468+171572.045}=\sqrt{226380.513}=475.795$